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Answers without the blur. Sign up and see all textbooks for free! Q. 6.35

Expert-verified Found in: Page 246 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Verify from Maxwell speed distribution that the most likely speed of a molecule is $\sqrt{2kT}{m}}$.

The most likely speed of a molecule is ${v}_{\mathrm{mp}}=\sqrt{2kT}{m}}$.

See the step by step solution

## Step 1. Given information

Maxwell speed distribution function is given by

$D\left(v\right)={\left(\frac{m}{2\pi kT}\right)}^{3}{2}}4\pi {v}^{2}{e}^{-m{v}^{2}}{2kT}}..........................\left(1\right)$

## Step 2. Calculation

Differentiate both sides of equation (1) with respect to $v$.

role="math" localid="1646971259346" $\begin{array}{rcl}\frac{dD\left(v\right)}{dv}& =& \frac{d}{dv}\left[{\left(\frac{m}{2\pi kT}\right)}^{3}{2}}4\pi {v}^{2}{e}^{-m{v}^{2}}{2kT}}\right]\\ & =& {\left(\frac{m}{2\pi kT}\right)}^{3}{2}}4\pi \left[2v{e}^{-m{v}^{2}}{2kT}}-\frac{2m{v}^{3}}{2kT}{e}^{-m{v}^{2}}{2kT}}\right]\\ & =& 8\pi {\left(\frac{m}{2\pi kT}\right)}^{3}{2}}v{e}^{-m{v}^{2}}{2kT}}\left[1-\frac{m{v}^{2}}{2kT}\right]............................\left(2\right)\end{array}$

At most likely speed of the gas molecules, the derivative of the Maxwell speed distribution function will be zero.

Substitute ${v}_{\mathrm{mp}}$ for $v$ and $0$ for $\begin{array}{rcl}& & \frac{dD\left(v\right)}{dv}\end{array}$ into equation (2) and simplify to obtain the most likely speed.

$\begin{array}{rcl}0& =& 8\pi {\left(\frac{m}{2\pi kT}\right)}^{3}{2}}v{e}^{-m{v}^{2}}{2kT}}\left[1-\frac{m{v}^{2}}{2kT}\right]\\ \left[1-\frac{m{{v}_{\mathrm{mp}}}^{2}}{2kT}\right]& =& 0\left[\mathrm{As}\mathrm{v}\ne 0\right]\\ {v}_{\mathrm{mp}}& =& \sqrt{\frac{2kT}{m}}\end{array}$

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